Reaction of Amines with Nitrous Acid (HNO2): Aliphatic vs Aromatic

Chemistry · Amines · NEET

Nitrous acid (HNO2, made fresh from NaNO2 + HCl) reacts with primary amines to make a diazonium salt. But there is a big difference: a primary aliphatic amine gives an unstable diazonium salt that breaks apart at once, releasing N2 gas and giving an alcohol. A primary aromatic amine (like aniline) gives a diazonium salt that is stable when kept cold at 273-278 K (0-5°C). Memory hook: "Aliphatic = Alcohol + bubbles; Aromatic = stays only if Arctic cold."
Reaction of 1° Amines with Nitrous Acid (HNO2 from NaNO2 + HCl)Aliphatic 1° amineR-NH2HNO2[ R-N2+ ] unstableR-OH + N2↑ + H2OAromatic 1° amineC6H5-NH2 (aniline)NaNO2/HCl, 273-278 KC6H5-N2+ Cl−stable ONLY when coldwarm → phenol + N2↑
Two paths with HNO2: a primary aliphatic amine gives an alcohol plus N2 gas at once (unstable diazonium), while a primary aromatic amine gives a diazonium salt that survives only at 273-278 K and turns to phenol if warmed.

Your doubts, answered

What is nitrous acid (HNO2) and how is it made in the reaction?

HNO2 is a weak, unstable acid. You cannot store it in a bottle. It is made fresh in the flask by mixing sodium nitrite (NaNO2) with a cold mineral acid like HCl: NaNO2 + HCl -> HNO2 + NaCl. This is why NEET writes the reagent as 'NaNO2/HCl' or 'NaNO2 + dil. HCl at 273-278 K'. They all mean the same thing: nitrous acid formed in place.

Why does a primary aliphatic amine give an alcohol instead of a stable salt?

A primary aliphatic amine (R-NH2) first forms an aliphatic diazonium salt (R-N2+). But this ion has no benzene ring to spread out and hold the positive charge, so it is very unstable. It breaks apart instantly, throwing out nitrogen gas (N2) and leaving a carbocation. Water then attacks to give an alcohol. Net result: R-NH2 + HNO2 -> R-OH + N2 (gas) + H2O. You see bubbles of N2. This is why aliphatic amines do NOT give a usable diazonium salt.

Why is the aromatic diazonium salt stable but only when cold?

In aniline, the -N2+ group is attached to a benzene ring. The ring lone pairs and pi electrons spread out (delocalise) the positive charge by resonance, so the salt is stabilised. But this stability only holds at low temperature, 273-278 K (0-5°C). If you warm it above about 278 K, it decomposes: the diazonium group is replaced by -OH and N2 escapes, giving phenol. So you must keep the reaction in an ice bath.

Do secondary and tertiary amines react with nitrous acid the same way?

No. Only PRIMARY amines give diazonium chemistry. A secondary amine (R2NH) reacts with HNO2 to give a yellow oily N-nitrosamine (R2N-N=O). A tertiary aliphatic amine just forms a soluble nitrite salt (no useful product). A tertiary aromatic amine like N,N-dimethylaniline gives a green C-nitroso ring product (p-nitroso). So the reaction with HNO2 can help tell the three classes apart, but the clean N2-gas/alcohol test is only for primary amines.

What is the difference between diazotisation and deamination here?

Diazotisation means making the diazonium salt (-N2+) from a primary amine using NaNO2/HCl at low temperature. Deamination means removing the nitrogen: for an aliphatic amine this happens on its own (N2 leaves, alcohol forms), and for an aromatic diazonium salt you can force it with warm water (gives phenol) or with H3PO2 (gives just the C-H, the amino group is fully removed). NEET loves asking which reagent removes the diazonium group.

Why does NEET always write the temperature 273-278 K?

Because temperature decides the product. At 273-278 K (0-5°C) aniline gives the stable benzenediazonium chloride you can use for further reactions (Sandmeyer, azo coupling). Above this range the same salt decomposes to phenol. So the number 273-278 K is a signal in a question that diazotisation is happening. If a statement says the aromatic salt is stable 'above 300 K', that is FALSE (this is a common NEET trap).

⚠️ The NEET trap
Primary aromatic amines react with HNO2 to form diazonium salts that are stable even above 300 K.
Aromatic diazonium salts are stable only at low temperature (273-278 K). Above that they decompose to phenol + N2. Only the aliphatic diazonium salt is called 'unstable' because it decomposes instantly at any temperature.
🧠 'Stable above 300 K' is a lie the exam plants. Aromatic diazonium is stable ONLY in the ice bath (273-278 K). Warm it = phenol + bubbles.

Real NEET questions

NEET 2022

Statement I: Primary aliphatic amines react with HNO2 to give unstable diazonium salts. Statement II: Primary aromatic amines react with HNO2 to form diazonium salts which are stable even above 300 K. Choose the most appropriate answer.

A · Both Statement I and Statement II are correct.
B · Both Statement I and Statement II are incorrect.
C · Statement I is correct but Statement II is incorrect.
D · Statement I is incorrect but Statement II is correct.
Solution: Statement I is correct: R-NH2 + HNO2 -> unstable aliphatic diazonium ion (R-N2+) that decomposes at once, releasing N2. Statement II is incorrect: aromatic (arene) diazonium salts are stabilised by ring resonance but only at LOW temperature, 273-278 K. Above about 278 K they decompose to phenol, so they are NOT stable above 300 K. Hence Statement I true, Statement II false = option (C).
NEET 2026

The major product Z formed in the following sequence of reactions is: C2H6 ->[Cl2/UV] X ->[NH3] Y ->[(i) NaNO2/HCl (ii) H2O] Z

A · C2H5NO2
B · C2H5-N=N-OH
C · C2H5OH
D · C2H5NH2
Solution: Step 1: free-radical monochlorination of ethane gives chloroethane, C2H6 -> C2H5Cl (X). Step 2: ammonolysis with NH3 gives the primary amine ethylamine, C2H5Cl -> C2H5NH2 (Y). Step 3: Y reacts with nitrous acid (NaNO2/HCl). Because the aliphatic diazonium ion C2H5N2+ has no ring to stabilise it, it breaks apart instantly, releasing N2, and with water gives the alcohol: C2H5NH2 + HNO2 -> C2H5OH + N2 + H2O. So Z = ethanol (C). This is the classic 'aliphatic amine + HNO2 = alcohol + N2' result.
NEET 2016 Phase 2

A nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2, to give an unstable compound B. B, on treatment with phenol, forms a coloured compound C with molecular formula C12H10N2O. The structure of compound A is:

A · Aniline, C6H5NH2
B · Nitrobenzene, C6H5NO2
C · Benzonitrile, C6H5CN
D · Benzamide, C6H5CONH2
Solution: The path is reduce -> diazotise -> couple. Sn/HCl reduces a nitro group to a primary aromatic amine, so A must be nitrobenzene: C6H5NO2 -> C6H5NH2. Then HNO2 (NaNO2/HCl at 273-278 K) diazotises aniline to the unstable benzenediazonium salt B, C6H5N2+Cl-. B couples with phenol to give the orange azo dye p-hydroxyazobenzene, C6H5-N=N-C6H4-OH = C12H10N2O = compound C. Working backward, A is nitrobenzene, option (B).

Solved Amines NEET PYQs

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Frequently asked

What is the main product of a primary aliphatic amine with HNO2?

An alcohol plus nitrogen gas. For example ethylamine gives ethanol: C2H5NH2 + HNO2 -> C2H5OH + N2 + H2O. The escaping N2 gas (bubbles) is the tell-tale sign of a primary aliphatic amine.

What is the product of aniline with HNO2?

Benzenediazonium chloride, C6H5N2+Cl-, when done cold at 273-278 K with NaNO2/HCl. This salt is useful for making many aromatic compounds through Sandmeyer, Gattermann and azo-coupling reactions.

Why must diazotisation of aniline be done at 0-5°C?

Because the benzenediazonium salt is only stable while cold (273-278 K). If the mixture warms above about 278 K, the salt decomposes to phenol and N2 gas, and your product is lost.

Can this reaction distinguish primary, secondary and tertiary amines?

Yes, partly. A primary amine gives N2 gas (aliphatic) or a diazonium salt (aromatic). A secondary amine gives a yellow oily nitrosamine. Tertiary amines give either a nitrite salt (aliphatic) or a nitroso ring compound (aromatic). But the sharpest single test that separates all three classes is the Hinsberg test.

Is HNO2 stored as a reagent?

No. Nitrous acid is unstable, so it is always prepared fresh in the flask by mixing sodium nitrite (NaNO2) with a cold mineral acid such as dilute HCl.